### Related Formula textComposite sequence relation: f^n(x) = fracx(1 + n x^4)^1/4$\text{Composite sequence relation: } f^n(x) = \frac{x}{(1 + n x^4)^{1/4}}$ ### Core Logic Evaluate the composition function step-by-step: f(f(x)) = fracf(x)(1 + f(x)^4)^1/4 = fracfracx(1+x^4)^1/4left(1 + fracx^41+x^4right)^1/4 = fracx(1 + 2x^4)^1/4$f(f(x)) = \frac{f(x)}{(1 + f(x)^4)^{1/4}} = \frac{\frac{x}{(1+x^4)^{1/4}}}{\left(1 + \frac{x^4}{1+x^4}\right)^{1/4}} = \frac{x}{(1 + 2x^4)^{1/4}}$ Continuing this pattern n$n$ times yields: f^n(x) = fracx(1 + nx^4)^1/4$f^n(x) = \frac{x}{(1 + nx^4)^{1/4}}$ Thus, g(x) = f(f(f(f(x)))) = f^4(x) = fracx(1 + 4x^4)^1/4$g(x) = f(f(f(f(x)))) = f^4(x) = \frac{x}{(1 + 4x^4)^{1/4}}$. ### Step 1: Setting up the Integral The required integral is: I = 18int_0^sqrt2sqrt5 fracx^3(1 + 4x^4)^1/4 dx$I = 18\int_{0}^{\sqrt{2\sqrt{5}}} \frac{x^3}{(1 + 4x^4)^{1/4}} dx$ We use substitution to solve this. Let 1 + 4x^4 = t^4$1 + 4x^4 = t^4$. Differentiating gives: 16x^3 dx = 4t^3 dt Rightarrow x^3 dx = frac14 t^3 dt$16x^3 dx = 4t^3 dt \Rightarrow x^3 dx = \frac{1}{4} t^3 dt$ ### Step 2: Modifying limits and Solving Change limits: When x = 0$x = 0$, 1 + 0 = t^4 Rightarrow t = 1$1 + 0 = t^4 \Rightarrow t = 1$. When x = (2sqrt5)^1/2$x = (2\sqrt{5})^{1/2}$, x^4 = (2sqrt5)^2 = 20$x^4 = (2\sqrt{5})^2 = 20$. Then 1 + 4(20) = 81 = t^4 Rightarrow t = 3$1 + 4(20) = 81 = t^4 \Rightarrow t = 3$. Substitute into the integral: I = 18 int_1^3 fracfrac14 t^3 dtt$I = 18 \int_{1}^{3} \frac{\frac{1}{4} t^3 dt}{t}$ I = frac184 int_1^3 t^2 dt = frac92 left[ fract^33 right]_1^3$I = \frac{18}{4} \int_{1}^{3} t^2 dt = \frac{9}{2} \left[ \frac{t^3}{3} \right]_1^3$ I = frac92 cdot frac13 [t^3]_1^3 = frac32 (27 - 1) = frac32 (26) = 39$I = \frac{9}{2} \cdot \frac{1}{3} [t^3]_1^3 = \frac{3}{2} (27 - 1) = \frac{3}{2} (26) = 39$ ### Pattern Recognition Repeated compositions of x(1+cx^k)^-1/k$x(1+cx^k)^{-1/k}$ follow a strict linearity in the coefficient of x^k$x^k$. Here, 4 compositions cleanly changed the x^4$x^4$ coefficient from 1 to 4, preparing a perfect variable substitution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Functions Class 12 Maths: Integral Calculus

### Related Formula textArea enclosed integrating w.r.t y-axis: int (x_textright - x_textleft) dy$\text{Area enclosed integrating w.r.t y-axis: } \int (x_{\text{right}} - x_{\text{left}}) dy$ ### Core Logic Find intersection points between the parabola x = (y-2)^2 + 1$x = (y-2)^2 + 1$ and the line x = 2y - 4$x = 2y - 4$. Set them equal: (y-2)^2 + 1 = 2y - 4$(y-2)^2 + 1 = 2y - 4$ y^2 - 4y + 4 + 1 = 2y - 4$y^2 - 4y + 4 + 1 = 2y - 4$ y^2 - 6y + 9 = 0 Rightarrow (y-3)^2 = 0$y^2 - 6y + 9 = 0 \Rightarrow (y-3)^2 = 0$ The line is tangent to the parabola at y = 3$y = 3$. At y = 3$y = 3$, x = 2(3) - 4 = 2$x = 2(3) - 4 = 2$. We need the area enclosed by the parabola, the line, and the *positive coordinate axes*. Let's trace the boundary intercepts. The line x = 2y - 4$x = 2y - 4$ cuts the y-axis (x=0$x=0$) at y=2$y=2$. Parabola vertex is at (1, 2)$(1, 2)$. Parabola cuts y-axis (x=0$x=0$) at (y-2)^2 = -1$(y-2)^2 = -1$ (No real y-intercept). So the curve (y-2)^2 = x-1$(y-2)^2 = x-1$ only exists for x ge 1$x \ge 1$. ### Step 1: Setting up the Integral w.r.t y

Because the area is bound by the positive coordinate axes, we track integration strictly in the first quadrant. The required area is the area between the y-axis, the x-axis, the line, and the curve. Wait, the problem says "area enclosed by the parabola... line... and positive coordinate axes". Let's integrate with respect to y$y$. The boundary curves are x = (y-2)^2 + 1$x = (y-2)^2 + 1$ (which forms the rightmost boundary) and the axes. The area bounded by the curve with y-axis is from y=0$y=0$ to y=3$y=3$ minus the triangular region cut by the line below y=2$y=2$. The line intersects the x-axis (y=0$y=0$) at x=-4$x=-4$ and y-axis (x=0$x=0$) at y=2$y=2$. The positive coordinate axes enforce boundaries at x ge 0, y ge 0$x \ge 0, y \ge 0$. So the exact area is the integral of the parabola's x$x$-value from y=0$y=0$ to y=3$y=3$, minus the area of the small triangle outside the region bounded by the line x=2y-4$x=2y-4$ and the axes in the positive quadrant. The area under the parabola (to the left towards the y-axis) from y=0$y=0$ to y=3$y=3$ is int_0^3 x_textparabola dy$\int_0^3 x_{\text{parabola}} dy$. The line bounds the region on the left from y=2$y=2$ to y=3$y=3$. Between y=0$y=0$ and y=2$y=2$, the area goes fully to the y-axis. So, area = int_0^3 x_textparabola dy - textArea of Delta text bounded by line and y-axis$\int_0^3 x_{\text{parabola}} dy - \text{Area of } \Delta \text{ bounded by line and y-axis}$. The triangle bounded by x=2y-4, x=0, y=2, y=3$x=2y-4, x=0, y=2, y=3$ is: base is x=2(3)-4 = 2$x=2(3)-4 = 2$ at y=3$y=3$, height is 1$1$ (from y=2$y=2$ to y=3$y=3$). Triangle area = frac12 times 2 times 1 = 1$\frac{1}{2} \times 2 \times 1 = 1$. Hence, Enclosed Area = int_0^3 ((y-2)^2 + 1) dy - 1$\int_{0}^{3} ((y-2)^2 + 1) dy - 1$. ### Step 2: Evaluating the Integral textArea = int_0^3 (y^2 - 4y + 5) dy - 1$\text{Area} = \int_{0}^{3} (y^2 - 4y + 5) dy - 1$ = left[ fracy^33 - 2y^2 + 5y right]_0^3 - 1$= \left[ \frac{y^3}{3} - 2y^2 + 5y \right]_0^3 - 1$ = left( frac273 - 2(9) + 5(3) right) - 0 - 1$= \left( \frac{27}{3} - 2(9) + 5(3) \right) - 0 - 1$ = (9 - 18 + 15) - 1 = 6 - 1 = 5$= (9 - 18 + 15) - 1 = 6 - 1 = 5$ ### Pattern Recognition Instead of evaluating horizontal strips with a broken left-bound, evaluate the total area mapped to the y-axis int_0^3 x dy$\int_0^3 x dy$ and simply subtract the basic geometric triangle removed by the straight line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integral Calculus

### Related Formula lim_n to infty frac1n sum_k=1^n fleft(fracknright) = int_0^1 f(x) dx$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n}\right) = \int_0^1 f(x) dx$ ### Core Logic Rewrite the series by extracting n^4$n^4$ from the denominator: L = lim_nto inftysum_k = 1^nfracn^3n^4left(1 + frack^2n^2right)left(1 + frac3k^2n^2right)$L = \lim_{n\to \infty}\sum_{k = 1}^{n}\frac{n^3}{n^4\left(1 + \frac{k^2}{n^2}\right)\left(1 + \frac{3k^2}{n^2}\right)}$ L = lim_n rightarrow infty frac1n sum_k = 1^n frac1left(1 + frack^2n^2right)left(1 + frac3k^2n^2right)$L = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{\left(1 + \frac{k^2}{n^2}\right)\left(1 + \frac{3k^2}{n^2}\right)}$ Using the limit of a sum, we convert this to a definite integral by substituting x = frackn$x = \frac{k}{n}$ and dx = frac1n$dx = \frac{1}{n}$: I = int_0^1 fracdx(1 + x^2)(1 + 3x^2)$I = \int_{0}^{1} \frac{dx}{(1 + x^2)(1 + 3x^2)}$ ### Step 1: Partial Fractions To integrate, we split the integrand using partial fractions. Notice that: frac1(1+x^2)(1+3x^2) = fracA1+x^2 + fracB1+3x^2$\frac{1}{(1+x^2)(1+3x^2)} = \frac{A}{1+x^2} + \frac{B}{1+3x^2}$ Let x^2 = t$x^2 = t$. Then frac1(1+t)(1+3t) = fracA1+t + fracB1+3t$\frac{1}{(1+t)(1+3t)} = \frac{A}{1+t} + \frac{B}{1+3t}$. 1 = A(1+3t) + B(1+t)$1 = A(1+3t) + B(1+t)$ If t = -1$t = -1$, 1 = A(-2) Rightarrow A = -frac12$1 = A(-2) \Rightarrow A = -\frac{1}{2}$. If t = -1/3$t = -1/3$, 1 = B(2/3) Rightarrow B = frac32$1 = B(2/3) \Rightarrow B = \frac{3}{2}$. So, frac1(1+x^2)(1+3x^2) = frac3/21+3x^2 - frac1/21+x^2$\frac{1}{(1+x^2)(1+3x^2)} = \frac{3/2}{1+3x^2} - \frac{1/2}{1+x^2}$ Alternatively, as formatted in the solution: I = int_0^1 fracdx3(1 + x^2)left(frac13 + x^2right)$I = \int_{0}^{1} \frac{dx}{3(1 + x^2)\left(\frac{1}{3} + x^2\right)}$ = int_0^1 frac13 times frac32 frac(x^2 + 1) - left(x^2 + frac13right)(1 + x^2)left(x^2 + frac13right) dx$= \int_{0}^{1} \frac{1}{3} \times \frac{3}{2} \frac{(x^2 + 1) - \left(x^2 + \frac{1}{3}\right)}{(1 + x^2)\left(x^2 + \frac{1}{3}\right)} dx$ = frac12 int_0^1 left[ frac1x^2 + left(frac1sqrt3right)^2 - frac11 + x^2 right] dx$= \frac{1}{2} \int_{0}^{1} \left[ \frac{1}{x^2 + \left(\frac{1}{\sqrt{3}}\right)^2} - \frac{1}{1 + x^2} \right] dx$ ### Step 2: Integration Integrating both terms: = frac12 left[ sqrt3 tan^-1 (sqrt3 x) right]_0^1 - frac12 left( tan^-1 x right)_0^1$= \frac{1}{2} \left[ \sqrt{3} \tan^{-1} (\sqrt{3} x) \right]_{0}^{1} - \frac{1}{2} \left( \tan^{-1} x \right)_{0}^{1}$ Evaluating the limits: = fracsqrt32 left(fracpi3right) - frac12 left(fracpi4right)$= \frac{\sqrt{3}}{2} \left(\frac{\pi}{3}\right) - \frac{1}{2} \left(\frac{\pi}{4}\right)$ = fracpi2 sqrt3 - fracpi8$= \frac{\pi}{2 \sqrt{3}} - \frac{\pi}{8}$ = frac4pi - sqrt3pi8sqrt3 = fracpi(4 - sqrt3)8sqrt3$= \frac{4\pi - \sqrt{3}\pi}{8\sqrt{3}} = \frac{\pi(4 - \sqrt{3})}{8\sqrt{3}}$ Rationalizing and finding the common denominator to match options: = frac13 pi8(4 sqrt3 + 3)$= \frac{13 \pi}{8(4 \sqrt{3} + 3)}$ (Note: Multiply numerator and denominator by (4sqrt3+3)$(4\sqrt{3}+3)$. Numerator becomes pi(4-sqrt3)(4sqrt3+3) = pi(16sqrt3 + 12 - 12 - 3sqrt3) = 13pisqrt3$\pi(4-\sqrt{3})(4\sqrt{3}+3) = \pi(16\sqrt{3} + 12 - 12 - 3\sqrt{3}) = 13\pi\sqrt{3}$. Wait, let's verify: fracpi2sqrt3 - fracpi8 = fracpi(4-sqrt3)8sqrt3$\frac{\pi}{2\sqrt{3}} - \frac{\pi}{8} = \frac{\pi(4-\sqrt{3})}{8\sqrt{3}}$. Let's rationalize the denominator in the given option frac13pi8(4sqrt3+3) = frac13pi(4sqrt3-3)8(48-9) = frac13pi(4sqrt3-3)8(39) = fracpi(4sqrt3-3)24 = frac4sqrt3pi24 - frac3pi24 = fracpi2sqrt3 - fracpi8$\frac{13\pi}{8(4\sqrt{3}+3)} = \frac{13\pi(4\sqrt{3}-3)}{8(48-9)} = \frac{13\pi(4\sqrt{3}-3)}{8(39)} = \frac{\pi(4\sqrt{3}-3)}{24} = \frac{4\sqrt{3}\pi}{24} - \frac{3\pi}{24} = \frac{\pi}{2\sqrt{3}} - \frac{\pi}{8}$. Match is exact.) ### Pattern Recognition Divide numerator and denominator by the highest power of n$n$ to convert the series into the standard form of Riemann Sum frac1n sum f(k/n)$\frac{1}{n} \sum f(k/n)$. Then use partial fractions for rational algebraic integrands. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integrals

### Related Formula textLeibnitz Rule: fracddxint_alpha(x)^beta(x) h(t)dt = h(beta(x))beta'(x) - h(alpha(x))alpha'(x)$\text{Leibnitz Rule: } \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} h(t)dt = h(\beta(x))\beta'(x) - h(\alpha(x))\alpha'(x)$ ### Core Logic For x>0$x>0$, f(x) = int_-x^x left( |t| - t^2 right) e^-t^2 dt$f(x) = \int_{-x}^{x} \left( |t| - t^2 \right) e^{-t^2} dt$. f'(x) = left(|x|-x^2right)e^-x^2 cdot 1 - left(|-x|-(-x)^2right)e^-(-x)^2 cdot (-1) = 2(x-x^2)e^-x^2$f'(x) = \left(|x|-x^2\right)e^{-x^2} \cdot 1 - \left(|-x|-(-x)^2\right)e^{-(-x)^2} \cdot (-1) = 2(x-x^2)e^{-x^2}$ For g(x) = int_0^x^2 t^frac12 e^-t dt$g(x) = \int_0^{x^2} t^{\frac{1}{2}} e^{-t} dt$: g'(x) = (x^2)^1/2 e^-x^2 cdot 2x = 2x^2 e^-x^2$g'(x) = (x^2)^{1/2} e^{-x^2} \cdot 2x = 2x^2 e^{-x^2}$ Sum of derivatives: f'(x) + g'(x) = 2xe^-x^2 - 2x^2e^-x^2 + 2x^2e^-x^2 = 2xe^-x^2$f'(x) + g'(x) = 2xe^{-x^2} - 2x^2e^{-x^2} + 2x^2e^{-x^2} = 2xe^{-x^2}$ Integrate both sides with respect to x$x$: f(x) + g(x) = int_0^x 2t e^-t^2 dt$f(x) + g(x) = \int_0^x 2t e^{-t^2} dt$ Let t^2 = u implies 2t dt = du$t^2 = u \implies 2t dt = du$: f(x) + g(x) = int_0^x^2 e^-u du = [-e^-u]_0^x^2 = 1 - e^-x^2$f(x) + g(x) = \int_0^{x^2} e^{-u} du = [-e^{-u}]_0^{x^2} = 1 - e^{-x^2}$ Substitute x = sqrtlog_e 9$x = \sqrt{\log_e 9}$: fleft(sqrtlog_e 9right) + gleft(sqrtlog_e 9right) = 1 - e^-log_e 9 = 1 - frac19 = frac89$f\left(\sqrt{\log_e 9}\right) + g\left(\sqrt{\log_e 9}\right) = 1 - e^{-\log_e 9} = 1 - \frac{1}{9} = \frac{8}{9}$ The question asks for 9(f+g)$9(f+g)$ based on typical JEE formatting of this problem (the PDF states "9(f(x)+g(x))$9(f(x)+g(x))$" in the solution line). So 9 times frac89 = 8$9 \times \frac{8}{9} = 8$. ### Pattern Recognition When dealing with definite integrals with variable limits, apply Leibnitz rule to convert to differential form, combine terms, and integrate back. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integrals